`HgCl_(2)` and `I_(2)` both when dissolved in water containing `I^(-)`ions, the pair of species formed is :

Options

A. `Hgl_(2),I^(-)`
B. `Hgl_(4)^(2-), I_(3)^(-)`
C. `Hg_(2)l_(2), I^(-)`
D. `Hgl_(2), I_(3)^(-)`

Explanation

To solve the problem, we need to analyze the reactions that occur when `HgCl2` and `I2` are dissolved in water containing `I^(-)` ions.

### Step-by-Step Solution:

1. **Identify the compounds involved**: We have `HgCl2` (mercury(II) chloride) and `I2` (iodine) in the presence of `I^(-)` ions.

2. **Reaction of `HgCl2` with `I^(-)` ions**:
- When `HgCl2` is added to a solution containing `I^(-)` ions, a complex ion is formed. The reaction can be represented as:
\[ HgCl_2 + 4I^- \rightarrow HgI_4^{2-} + 2Cl^- \]
- Here, `HgI4^{2-}` (tetraiodomercurate(II)) is formed along with `Cl^-` ions.

3. **Reaction of `I2` with `I^(-)` ions**:
- When `I2` is added to the solution containing `I^(-)` ions, another species is formed. The reaction can be represented as:
\[ I_2 + I^- \rightarrow I_3^- \]
- In this reaction, the triiodide ion `I3^-` is formed.

4. **Summary of the species formed**:
- From the reactions above, we have two species formed:
- `HgI4^{2-}`
- `I3^-`

5. **Conclusion**:
- The pair of species formed when `HgCl2` and `I2` are dissolved in water containing `I^(-)` ions is `HgI4^{2-}` and `I3^-`.

### Final Answer:
The pair of species formed is `HgI4^{2-}` and `I3^-`.

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